题目：

分析：

线段树

代码：

// Problem: F - Two Sequence Queries
// Contest: AtCoder - SuntoryProgrammingContest2024（AtCoder Beginner Contest 357）
// URL: https://atcoder.jp/contests/abc357/tasks/abc357_f
// Memory Limit: 1024 MB
// Time Limit: 5000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include<bits/stdc++.h>

using namespace std;
#define int long long

const int N = 2e6+5;
const int mod=998244353;

int n,q;
int a[N],b[N];
int taga[N],tagb[N];
int sum[N],suma[N],sumb[N];

void addtag(int p,int pl,int pr,int d,int k){
	if(k==1){
		(suma[p]+=1ll*d*(pr-pl+1)%mod)%=mod;
		(sum[p]+=1ll*sumb[p]*d%mod)%=mod;
		(taga[p]+=d)%=mod;
	}
	else{
		(sumb[p]+=1ll*d*(pr-pl+1)%mod)%=mod;
		(sum[p]+=1ll*suma[p]*d%mod)%=mod;
		(tagb[p]+=d)%=mod;
	}
}

void pushup(int p){
	(suma[p]=(suma[p*2]+suma[p*2+1])%mod)%=mod;
	(sumb[p]=(sumb[p*2]+sumb[p*2+1])%mod)%=mod;
	(sum[p]=(sum[p*2]+sum[p*2+1])%mod)%=mod;
}

void pushdown(int p,int pl,int pr){
	int mid=(pl+pr)/2;
	if(taga[p]){
		addtag(p*2,pl,mid,taga[p],1);
		addtag(p*2+1,mid+1,pr,taga[p],1);
		taga[p]=0;
	}
	if(tagb[p]){
		addtag(p*2,pl,mid,tagb[p],2);
		addtag(p*2+1,mid+1,pr,tagb[p],2);
		tagb[p]=0;
	}
}

void build(int p,int pl,int pr){
	if(pl==pr){
		suma[p]=a[pl]%mod;
		sumb[p]=b[pl]%mod;
		sum[p]=1ll*a[pl]*b[pl]%mod;
		taga[p]=tagb[p]=0;
		return;
	}
	int mid=(pl+pr)/2;
	build(p*2,pl,mid);
	build(p*2+1,mid+1,pr);
	pushup(p);
	return;
}

void update(int l,int r,int p,int pl,int pr,int d,int k){
	if(l<=pl&&pr<=r){
		addtag(p,pl,pr,d,k);
		return;
	}
	int mid=(pl+pr)/2;
	pushdown(p,pl,pr);
	if(l<=mid){
		update(l,r,p*2,pl,mid,d,k);
	}
	if(r>mid){
		update(l,r,p*2+1,mid+1,pr,d,k);
	}
	pushup(p);
	return;
}

int query(int l,int r,int p,int pl,int pr){
	if(l<=pl&&pr<=r){
		return sum[p];
	}
	int mid=(pl+pr)/2;
	pushdown(p,pl,pr);
	int ans=0;
	if(l<=mid){
		(ans+=query(l,r,p*2,pl,mid))%=mod;
	}
	if(r>mid){
		(ans+=query(l,r,p*2+1,mid+1,pr))%=mod;
	}
	return ans;
}

signed main(){
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	
	cin>>n>>q;
	for(int i=1;i<=n;i++){
		cin>>a[i];
	}
	for(int i=1;i<=n;i++){
		cin>>b[i];
	}
	build(1,1,n);
	
	while(q--){
		int op,l,r,x;
		cin>>op;
		if(op==1){
			cin>>l>>r>>x;
			update(l,r,1,1,n,x,1);
		}
		else if(op==2){
			cin>>l>>r>>x;
			update(l,r,1,1,n,x,2);
		}
		else if(op==3){
			cin>>l>>r;
			cout<<query(l,r,1,1,n)<<"\n";
		}
	}

	
	return 0;	

}